Spectral Decomposition

The spectral decomposition of a finite-dimensional Hermitian operator separates its possible spectral values from the orthogonal subspaces on which those values act. If $A=A^\dagger$, then

$$A=\sum_{a\in\sigma(A)}aP_a,$$

where the sum is over distinct eigenvalues and $P_a$ is the orthogonal projector onto the full eigenspace belonging to $a$.

This page is the mathematical home for the finite-dimensional theorem, including degeneracy and projector reconstruction. The interpretation of $a$ as a possible observable value and $P_a$ as an outcome sector belongs to Spectral Decomposition in Core Formalism.

Finite-Dimensional Spectral Theorem

Let $\mathcal H$ be a finite-dimensional complex Hilbert space and let $A:\mathcal H\to\mathcal H$ be Hermitian. Write

$$E_a=\ker(A-aI)$$

for the eigenspace associated with the distinct eigenvalue $a$. Then:

1. every $a\in\sigma(A)$ is real;
2. eigenspaces belonging to distinct eigenvalues are orthogonal;
3. the eigenspaces span the whole space;
4. there is an orthonormal basis consisting of eigenvectors of $A$.

Equivalently,

$$\mathcal H =\mathop{\bigoplus_{\!a\in\sigma(A)}}_{\perp}E_a.$$

If $P_a$ denotes the orthogonal projector onto $E_a$, the theorem takes the basis-independent form

$$\begin{aligned} A &=\sum_{a\in\sigma(A)}aP_a,\\ I &=\sum_{a\in\sigma(A)}P_a. \end{aligned}$$

The second identity is completeness. The first says that $A$ acts as scalar multiplication by $a$ on the subspace $E_a$.

Why the Theorem Holds

The proof can be organized as an induction on $\dim\mathcal H$.

Because the scalar field is $\mathbb C$, the characteristic polynomial has a root, so $A$ has an eigenvector $u$. Hermiticity makes its eigenvalue real. Normalize $u$ and write $Au=au$.

The orthogonal complement $u^\perp$ is invariant under $A$. If $v\in u^\perp$, then

$$\begin{aligned} \langle u\vert Av\rangle &=\langle Au\vert v\rangle\\ &=a\langle u\vert v\rangle\\ &=0. \end{aligned}$$

Thus $Av\in u^\perp$. The restriction of $A$ to $u^\perp$ remains Hermitian, so the same argument applies on a space of dimension one less. Induction produces an orthonormal eigenbasis of the whole space.

Orthogonality between distinct eigenspaces follows directly. If $Au=au$, $Av=bv$, and $a\ne b$, then

$$\begin{aligned} b\langle u\vert v\rangle &=\langle u\vert Av\rangle\\ &=\langle Au\vert v\rangle\\ &=a\langle u\vert v\rangle, \end{aligned}$$

so $\langle u\vert v\rangle=0$.

The proof uses finite dimensionality twice: an eigenvector exists, and the induction terminates. Infinite-dimensional self-adjoint operators require a different statement.

Degenerate Eigenspaces

Suppose the distinct eigenvalues are $a_1,\ldots,a_s$, and let

$$d_a=\dim E_a$$

be the multiplicity of $a$. Choose an orthonormal basis $\{\lvert a,\alpha\rangle\}_{\alpha=1}^{d_a}$ inside each eigenspace. Then

$$P_a =\sum_{\alpha=1}^{d_a} \lvert a,\alpha\rangle \langle a,\alpha\rvert$$

and

$$A =\sum_a\sum_{\alpha=1}^{d_a} a\lvert a,\alpha\rangle \langle a,\alpha\rvert.$$

The vectors inside a degenerate eigenspace are not unique. Any unitary change of basis within $E_a$ gives another orthonormal eigenbasis. The projector $P_a$, however, is unchanged. It is the canonical object associated with the degenerate spectral value.

For a Hermitian operator, algebraic and geometric multiplicities agree. The characteristic polynomial can therefore be written

$$\begin{aligned} \chi_A(z) &=\prod_{a\in\sigma(A)} (z-a)^{d_a},\\ \sum_a d_a &=\dim\mathcal H. \end{aligned}$$

The minimal polynomial has no repeated factor:

$$m_A(z) =\prod_{a\in\sigma(A)}(z-a).$$

This absence of repeated roots is the algebraic signature of diagonalizability.

Projector Algebra

The spectral projectors form an orthogonal resolution of the identity:

$$\begin{aligned} P_a^\dagger &=P_a,\\ P_aP_b &=\delta_{ab}P_a,\\ \sum_aP_a &=I. \end{aligned}$$

They also satisfy

$$\begin{aligned} AP_a &=P_aA=aP_a,\\ \operatorname{Ran}P_a &=E_a. \end{aligned}$$

Conversely, suppose $\{P_a\}$ is any finite orthogonal resolution of the identity and the labels $a$ are real. Then

$$A=\sum_a aP_a$$

defines a Hermitian operator whose eigenspace for $a$ is $\operatorname{Ran}P_a$, provided distinct projectors carry distinct labels. The decomposition is therefore both a consequence and a characterization of finite-dimensional Hermiticity.

Recovering Projectors from the Operator

The eigenspace projectors can be recovered without choosing eigenvector phases or a basis inside a degenerate block. For each distinct eigenvalue $a$, define the Lagrange polynomial

$$p_a(z) =\prod_{\substack{ b\in\sigma(A)\\ b\ne a }} \frac{z-b}{a-b}.$$

It obeys $p_a(a)=1$ and $p_a(b)=0$ for every other spectral value $b$. Applying the polynomial to $A$ gives

$$P_a=p_a(A) =\prod_{\substack{ b\in\sigma(A)\\ b\ne a }} \frac{A-bI}{a-b}.$$

This identity proves directly that the spectral projectors are unique. It also shows that every operator commuting with $A$ commutes with each $P_a$.

For an operator with only two distinct eigenvalues $a$ and $b$,

$$\begin{aligned} P_a &=\frac{A-bI}{a-b},\\ P_b &=\frac{A-aI}{b-a}. \end{aligned}$$

If $A$ has just one eigenvalue, Hermiticity forces $A=aI$ and the single spectral projector is $I$.

Spectral Decomposition and Diagonalization

Choose an orthonormal eigenbasis and place its vectors in the columns of a unitary matrix $U$. If $D$ is the diagonal matrix of eigenvalues, repeated according to multiplicity, then

$$\begin{aligned} U^\dagger AU &=D,\\ A &=UDU^\dagger. \end{aligned}$$

This is unitary diagonalization. The diagonal matrix depends on an ordering, and $U$ is nonunique whenever phases or degenerate basis rotations are changed. By contrast,

$$A=\sum_a aP_a$$

is basis independent and groups all copies of one eigenvalue into a single canonical subspace.

General diagonalization may use a nonunitary similarity $A=SDS^{-1}$. The spectral theorem says that Hermitian, and more generally normal, matrices admit the stronger choice $S=U$ with $U^{-1}=U^\dagger$. See Diagonalization and Normal Operators.

Worked Example: A Degenerate Spectrum

Consider

$$A= \begin{pmatrix} \frac52&\frac32&0\\ \frac32&\frac52&0\\ 0&0&1 \end{pmatrix}.$$

The normalized vector

$$u=\frac{1}{\sqrt2} \begin{pmatrix} 1\\1\\0 \end{pmatrix}$$

has eigenvalue $4$. Its orthogonal complement is the two-dimensional eigenspace with eigenvalue $1$. The projectors are

$$\begin{aligned} P_4 &=\frac12 \begin{pmatrix} 1&1&0\\ 1&1&0\\ 0&0&0 \end{pmatrix},\\[4pt] P_1 &=\frac12 \begin{pmatrix} 1&-1&0\\ -1&1&0\\ 0&0&2 \end{pmatrix}. \end{aligned}$$

They satisfy

$$P_4P_1=0, \qquad P_4+P_1=I,$$

and the spectral decomposition is

$$A=4P_4+P_1.$$

The polynomial formulas recover the same projectors:

$$\begin{aligned} P_4 &=\frac{A-I}{3},\\ P_1 &=\frac{4I-A}{3}. \end{aligned}$$

An orthonormal basis of the eigenvalue-$1$ subspace could be $(1,-1,0)^T/\sqrt2$ and $(0,0,1)^T$, but rotating those two vectors leaves $P_1$ and $A$ unchanged.

Information Read from the Spectrum

Once the distinct values $a$ and multiplicities $d_a$ are known, many operator properties are immediate:

$$\begin{aligned} \operatorname{tr}A &=\sum_a d_a a,\\ \det A &=\prod_a a^{d_a},\\ \operatorname{rank}A &=\sum_{a\ne0}d_a,\\ \lVert A\rVert_{\mathrm{op}} &=\max_a\lvert a\rvert. \end{aligned}$$

If $0\in\sigma(A)$, then

$$\ker A=\operatorname{Ran}P_0.$$

The operator is positive semidefinite precisely when every $a\ge0$, and it is positive definite precisely when every $a>0$.

For a normalized vector $\psi$,

$$\langle\psi\vert A\vert\psi\rangle =\sum_a a\lVert P_a\psi\rVert^2.$$

Because the nonnegative weights sum to one, the quadratic form lies between the smallest and largest eigenvalues. This is the spectral origin of the Rayleigh bounds.

Powers and Functions

Orthogonality eliminates mixed products. For every positive integer $n$,

$$A^n=\sum_a a^nP_a.$$

More generally, for any function defined on the finite set $\sigma(A)$,

$$f(A)=\sum_a f(a)P_a.$$

Examples include

$$\begin{aligned} e^A &=\sum_a e^aP_a,\\ A^{-1} &=\sum_a\frac{1}{a}P_a \quad\text{if }0\notin\sigma(A). \end{aligned}$$

In finite dimensions, every such $f(A)$ can also be represented by a polynomial in $A$ through interpolation. The construction, analytic definitions, square roots, and exponentials are developed in Matrix Functions and Exponentials.

Commuting Hermitian Operators

Suppose $A$ and $B$ are Hermitian and $[A,B]=0$. Since each $P_a$ is a polynomial in $A$,

$$[B,P_a]=0.$$

Therefore $B$ preserves every eigenspace of $A$:

$$B(E_a)\subseteq E_a.$$

Within each possibly degenerate $E_a$, the restriction of $B$ is Hermitian and can be diagonalized using an orthonormal basis. Combining the bases from all blocks gives a common orthonormal eigenbasis for $A$ and $B$.

Thus two finite-dimensional Hermitian operators commute if and only if they are simultaneously unitarily diagonalizable. Degeneracy is not an obstacle; it is the freedom that allows the second operator to be diagonalized within the first operator’s spectral blocks.

Boundary with Quantum Measurement

The spectral theorem supplies values $a$ and orthogonal projectors $P_a$. It does not, by itself, supply the Born rule, identify a laboratory observable, or prescribe a post-measurement state. Those are physical inputs to quantum theory.

For the probabilistic interpretation, degeneracy, moments, coarse graining, and outcome sectors, see Spectral Decomposition in Core Formalism and The Discrete Born Rule. The mathematical decomposition should not be mistaken for a complete measurement model.

Infinite-Dimensional Qualification

The finite sum over eigenspaces is not the general spectral theorem. A self-adjoint operator may have continuous spectrum and no normalizable eigenvectors at some spectral values. The replacement is a projection-valued measure $E$:

$$A=\int_{\mathbb R}\lambda\,dE(\lambda).$$

For an unbounded operator, the integral also determines the operator domain; formal manipulation without domain control can be invalid.

Compact self-adjoint operators retain a largely discrete picture: their nonzero spectral values are eigenvalues of finite multiplicity, with no nonzero accumulation point. General self-adjoint operators need the full measure-theoretic formulation. See Spectral Theorem, Practical Version for that extension.

Numerical Practice

For a Hermitian matrix, use a Hermitian eigensolver. It returns an approximately unitary eigenvector matrix $Q$ and a real diagonal matrix $\Lambda$ such that

$$A\approx Q\Lambda Q^\dagger.$$

Useful diagnostics include

$$\begin{aligned} r_j &=\lVert Aq_j-\lambda_jq_j\rVert,\\ \epsilon_{\mathrm{orth}} &=\lVert Q^\dagger Q-I\rVert,\\ \epsilon_{\mathrm{rec}} &=\lVert A-Q\Lambda Q^\dagger\rVert. \end{aligned}$$

Report these relative to an appropriate scale such as $\lVert A\rVert$.

Individual eigenvectors inside an exactly or nearly degenerate cluster are not stable objects: tiny perturbations can rotate them substantially. The invariant subspace is the meaningful result. If $Q_C$ contains an orthonormal basis for a resolved eigenvalue cluster $C$, form

$$P_C=Q_CQ_C^\dagger.$$

Compare projectors or subspaces across calculations rather than comparing eigenvectors column by column. Clustering requires a tolerance informed by residuals, spectral gaps, matrix scale, and the underlying model, not merely a fixed number of decimal places. See Matrix Diagonalization for algorithms and conditioning.

Common Mistakes

• Summing over eigenvectors without grouping a degenerate eigenspace into one projector.
• Treating a chosen eigenbasis inside a degenerate subspace as canonical.
• Assuming a matrix with real eigenvalues must be Hermitian.
• Confusing general similarity diagonalization with unitary diagonalization.
• Applying a scalar function separately to matrix entries.
• Dividing by $a-b$ in a projector formula before checking that the spectral values are distinct.
• Comparing numerical eigenvectors directly inside a degenerate cluster.
• Replacing the infinite-dimensional spectral theorem by an unjustified sum over generalized eigenvectors.
• Inferring the Born rule or state-update law from linear algebra alone.

Exercises

1. Suppose

   $$A=\sum_a aP_a$$

   is a finite-dimensional spectral decomposition. Prove that $p(A)=\sum_a p(a)P_a$ for every polynomial $p$. Deduce the formula for $A^{-1}$ when no $a$ is zero.

Solution

The orthogonal projector algebra gives

$$P_aP_b=\delta_{ab}P_a.$$

Therefore, for every positive integer $n$,

$$A^n =\left(\sum_a aP_a\right)^n =\sum_a a^nP_a.$$

Linearity then gives

$$p(A)=\sum_a p(a)P_a$$

for any polynomial. If every $a\ne0$, define

$$B=\sum_a\frac{1}{a}P_a.$$

Then

$$AB=BA=\sum_aP_a=I,$$

so $B=A^{-1}$.

2. Let

   $$A= \begin{pmatrix} 2&i\\ -i&2 \end{pmatrix}.$$

   Find the distinct eigenvalues and recover their projectors as polynomials in $A$. Verify the spectral decomposition.

Solution

The characteristic polynomial is

$$\det(\lambda I-A) =(\lambda-2)^2-1,$$

so the eigenvalues are $1$ and $3$. The two-value formulas give

$$\begin{aligned} P_3 &=\frac{A-I}{2} =\frac12 \begin{pmatrix} 1&i\\ -i&1 \end{pmatrix},\\[4pt] P_1 &=\frac{3I-A}{2} =\frac12 \begin{pmatrix} 1&-i\\ i&1 \end{pmatrix}. \end{aligned}$$

Direct multiplication shows

$$P_1^2=P_1, \qquad P_3^2=P_3, \qquad P_1P_3=0.$$

Also $P_1+P_3=I$, and

$$P_1+3P_3 = \begin{pmatrix} 2&i\\ -i&2 \end{pmatrix} =A.$$

3. Let $a_1,\ldots,a_s$ be the distinct eigenvalues of a Hermitian operator $A$. Prove directly that

   $$P_j =\prod_{k\ne j} \frac{A-a_kI}{a_j-a_k}$$

   acts as the identity on $E_{a_j}$ and as zero on every other eigenspace.

Solution

Take $v\in E_{a_\ell}$, so $Av=a_\ell v$. Every factor in the product acts on $v$ by scalar multiplication:

$$P_jv = \left[ \prod_{k\ne j} \frac{a_\ell-a_k}{a_j-a_k} \right]v.$$

If $\ell=j$, every numerator equals its denominator, so $P_jv=v$. If $\ell\ne j$, the factor with $k=\ell$ has zero numerator, so $P_jv=0$. The eigenspaces span the Hilbert space, hence this operator is exactly the orthogonal projector onto $E_{a_j}$.

4. Let $A$ and $B$ be commuting Hermitian matrices. Show that every eigenspace of $A$ is invariant under $B$, and use this fact to construct a common orthonormal eigenbasis.

Solution

If $v\in E_a$, then $Av=av$. Commutation gives

$$A(Bv) =B(Av) =aBv.$$

Thus $Bv\in E_a$, so each eigenspace of $A$ is invariant under $B$. The restriction of $B$ to $E_a$ is Hermitian: for $u,v\in E_a$,

$$\langle u\vert Bv\rangle =\langle Bu\vert v\rangle.$$

Apply the finite-dimensional spectral theorem to this restriction and choose an orthonormal basis of $B$-eigenvectors within every $E_a$. Eigenspaces of $A$ are mutually orthogonal, so the union of these blockwise bases is an orthonormal basis of the full space. Every basis vector is an eigenvector of both $A$ and $B$.

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• R. A. Horn and C. R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press, 2013.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• L. N. Trefethen and D. Bau III, Numerical Linear Algebra, SIAM, 1997.